Using the null hypothesis on t - statistic, we get
, μ1 − μ2 ≠ 0 and μ₁>μ₂ , that is true mean number of sales at location a is exceds than the true mean number of sales at location b at the 0. 1 level of significance.
We have given that
For Location A ,
sample size (n₁) = 18
standard deviations (s₁) = 9
sample mean (m₁) = 39
For Location B,
sample size (n₂) = 18
standard deviations (s₂) = 6
sample mean (m₂) = 55
0.01 level of significance implies confidence interval 1% .
1% Confidence interval for μ₁ - μ₂.
A t-test is used when looking at a numeric variable (such as height) and comparing the means of two different populations or groups.
Null Hypothesis
H0: μ₁ -μ₂= 0, where u1 is the first population mean and u2 is her second population mean.
As above, the null hypothesis tends to be no difference between the two population means. Or, more formally, that the difference is zero.
Equation , t =( m₁ - m₂ )/ (√(n₁ -1)s²₁ + (n₂- 1)s²₂)/(n₂ + n₁ - 2)( 1/n₁ + 1/n₂)
=> t = -16 /(√(17×81 + 17×36)/32 )(1/18 + 1/18)
=> t = - 16 /2.627
=> t = - 6.090
so, basis of t -value the null hypothesis is rejected.
hence , (μ₁ − μ₂ )not equal to 0 and t = - 6.090
t = (m₁ - μ₁ )/ s₁
=> -6.090 ×9 = 39 - μ₁
=>μ ₁ = 93.1
similarly, μ₂ = -6.090 ×6 + 55 = 91. 54
=> μ₁>μ₂
so, true mean number of sales at location a exceeds the true mean number of sales at location b.
To learn more about Null hypothesis on t - statistic, refer:
https://brainly.com/question/14016208
#SPJ4